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Formatted question description: https://leetcode.ca/all/2319.html

2319. Check if Matrix Is X-Matrix

• Difficulty: Easy.
• Related Topics: Array, Matrix.
• Similar Questions: Matrix Diagonal Sum.

Problem

A square matrix is said to be an X-Matrix if both of the following conditions hold:

• All the elements in the diagonals of the matrix are non-zero.

• All other elements are 0.

Given a 2D integer array grid of size n x n representing a square matrix, return true** if grid is an X-Matrix**. Otherwise, return false.

  Example 1:

Input: grid = [[2,0,0,1],[0,3,1,0],[0,5,2,0],[4,0,0,2]]
Output: true
Explanation: Refer to the diagram above. 
An X-Matrix should have the green elements (diagonals) be non-zero and the red elements be 0.
Thus, grid is an X-Matrix.

Example 2:

Input: grid = [[5,7,0],[0,3,1],[0,5,0]]
Output: false
Explanation: Refer to the diagram above.
An X-Matrix should have the green elements (diagonals) be non-zero and the red elements be 0.
Thus, grid is not an X-Matrix.

  Constraints:

• n == grid.length == grid[i].length

• 3 <= n <= 100

• 0 <= grid[i][j] <= 105

Solution (Java, C++, Python)

• Java
• Python
• C++
• Go
• TypeScript
• C#
• RenderScript

• class Solution {
      public boolean checkXMatrix(int[][] grid) {
          for (int i = 0; i < grid.length; i++) {
              for (int j = 0; j < grid[0].length; j++) {
                  if (i == j || i + j == grid.length - 1) {
                      if (grid[i][j] == 0) {
                          return false;
                      }
                  } else {
                      if (grid[i][j] != 0) {
                          return false;
                      }
                  }
              }
          }
          return true;
      }
  }
  
  ############
  
  class Solution {
      public boolean checkXMatrix(int[][] grid) {
          int n = grid.length;
          for (int i = 0; i < n; ++i) {
              for (int j = 0; j < n; ++j) {
                  if (i == j || i + j == n - 1) {
                      if (grid[i][j] == 0) {
                          return false;
                      }
                  } else if (grid[i][j] != 0) {
                      return false;
                  }
              }
          }
          return true;
      }
  }
  

• class Solution:
      def checkXMatrix(self, grid: List[List[int]]) -> bool:
          for i, row in enumerate(grid):
              for j, v in enumerate(row):
                  if i == j or i + j == len(grid) - 1:
                      if v == 0:
                          return False
                  elif v:
                      return False
          return True
  
  ############
  
  # 2319. Check if Matrix Is X-Matrix
  # https://leetcode.com/problems/check-if-matrix-is-x-matrix
  
  class Solution:
      def checkXMatrix(self, grid: List[List[int]]) -> bool:
          rows, cols = len(grid), len(grid[0])
          
          for i in range(rows):
              for j in range(cols):
                  if i == j or i + j == cols - 1:
                      if grid[i][j] == 0:
                          return False
                  elif grid[i][j] != 0:
                          return False
                      
          return True
  
  

• class Solution {
  public:
      bool checkXMatrix(vector<vector<int>>& grid) {
          int n = grid.size();
          for (int i = 0; i < n; ++i) {
              for (int j = 0; j < n; ++j) {
                  if (i == j || i + j == n - 1) {
                      if (!grid[i][j]) {
                          return false;
                      }
                  } else if (grid[i][j]) {
                      return false;
                  }
              }
          }
          return true;
      }
  };
  

• func checkXMatrix(grid [][]int) bool {
  	for i, row := range grid {
  		for j, v := range row {
  			if i == j || i+j == len(row)-1 {
  				if v == 0 {
  					return false
  				}
  			} else if v != 0 {
  				return false
  			}
  		}
  	}
  	return true
  }
  

• function checkXMatrix(grid: number[][]): boolean {
      const n = grid.length;
      for (let i = 0; i < n; ++i) {
          for (let j = 0; j < n; ++j) {
              if (i == j || i + j == n - 1) {
                  if (!grid[i][j]) {
                      return false;
                  }
              } else if (grid[i][j]) {
                  return false;
              }
          }
      }
      return true;
  }
  
  

• public class Solution {
      public bool CheckXMatrix(int[][] grid) {
          int n = grid.Length;
          for (int i = 0; i < n; ++i) {
              for (int j = 0; j < n; ++j) {
                  if (i == j || i + j == n - 1) {
                      if (grid[i][j] == 0) {
                          return false;
                      }
                  } else if (grid[i][j] != 0) {
                      return false;
                  }
              }
          }
          return true;
      }
  }
  

• impl Solution {
      pub fn check_x_matrix(grid: Vec<Vec<i32>>) -> bool {
          let n = grid.len();
          for i in 0..n {
              for j in 0..n {
                  if i == j || i + j == n - 1 {
                      if grid[i][j] == 0 {
                          return false;
                      }
                  } else if grid[i][j] != 0 {
                      return false;
                  }
              }
          }
          true
      }
  }
  
  

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).

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